If 0.44 g of substance dissolved in 22.2 g of benzene lowers the freezing point of benzene by 0.567°C, then the molecular mass of a substance is, (the molal depression constant = 5.12°C mol^–1 kg)

178.9

The correct answer is option 3. 178.9.

Given,

Weight of substance \(= 0.440 g\)

Weight of substance \(= 22.2 g\)

\(\Delta T_f = 0.567^o\text{C kg mol}^{-1}\)

\(K_f = 5.12^oC\)

We know,

Depression in freezing point,

\(\Delta T_f = K_f × m\)

\(⇒ 0.567^oC = 5.12^o\text{C kg mol}^{-1} × \frac{0.44 × 1000}{M × 22.2 g}\)

\(⇒ M = \frac{5.12 × 0.44 ×1000}{22.2 × 0.567}\)

\(⇒ M = \frac{2252.8}{12.58}\)

or, \(M = 178.9\)